welcome to digital electronics lecture series I professor it is Dholakia is going to explain you boolean algebra examples in this session so here over this side you can see William algebra rules which I have already derived it in my previous videos right you can go through it and hear those rules that we will directly use it over here in examples right so here I'll give you first question so this is my first question right now see in this question if you observe we have a dot B over here and here we have a dot B whole bar right so if you consider a dot B is X then we can say this is X bar plus a bar plus X where here we have considered X is equals to a dot B right so X plus X bar you can see as per the rule X plus X bar right that is 1 X plus X bar that is 1 so we can say this is a bar plus 1 right and a bar plus 1 anything plus 1 that is 1 so we can say this is one's complement and once complement is 0 so this is how we can solve this problem now sometimes in examples you may not say this kind of thing like a B and a B bar is there and you can consider X so sometimes you may not see this so how to solve this so by having simple method even we can do this like see we can apply de Morgan's theorem here here right and as if you apply de Morgan's theorem further simplification can be easily done so let us solve this by having one more way so as if you apply de Morgan's theorem over here then all we need to do is we need to consider this as a variable right so plus will get translated into dot and dot will get translated into plus so here a plus sign is there so we need to translate that into dot and this variables complement that we need to apply over here so this will be a B double bar this will be a double bar and this will be a B bar right and complement of complement that is that variable only as per not rule you can see so here we can say this is a B dot ay ay B bar now see here a dot a that is a that we can say so we can say this is a dot ay means a so we can say this is a B dot ay B bar now C X dot X bar that is zero so we can say a B is equals to X right so that results in two X dot X bar is equals to zero so that is how we can solve this type of problem right now let us have second question so now in second question we have been given with so here we are deal with to solve this right by applying boolean algebra rules so as we know the priorities first we need to apply priority to complement and that is what we can do it over here by applying boolean algebra de Morgan's law right as if I apply boolean algebra de Morgan's law over here then this will be now you see over here we need to apply de Morgan's law so this will be now a me complement and plus will be now dot and AC bar complemented right so that is how we can apply de Morgan's law so again I'll apply de Morgan's law to a B bar and a dot C bar whole bar so now this will be a dot B whole bar that is a bar plus B bar and this dot into a plus C whole bar a DS a bar plus C double bar now here we can simplify this further we're here C double bar that is C right now to elaborate this equation now we will multiply these two terms so over here a bar dot a bar plus a bar dot C plus B bar dot a bar plus B bar dot C that we can say right now see here a bar dot a bar that is a bar only right as per ending rule you can see over here and let us simplify this further now see here in this three terms in this three terms I can take a bar common you can see right and as if you take a bar common from these three terms you will be finding like see here there will be one here there will be plus C then plus here B bar right so that will solve this equation further so c1 plus anything that is one only right so with this term if you apply one plus anything that is equals to one we can say this is one only and one dot a bar that is a bar only so simplified equation will be now we should do further multiplication of this term so now we will multiply this term with this term so that is C bar a bar plus B bar C C bar right now here you can see there is C C bar and C C bar is 0 so this term will get cancelled now so this will be a dot B plus C bar a bar now we can multiply this term so this will be a B plus C bar a bar a now here you can see this is a a bar so a a bar that will get cancelled that will be 0 as per ending rule so we can see this is now a B right so by using demorgan's theorum and basic rules of boolean algebra we can be able to solve this type of question right let us have one more question so I am going to write one more question over here so the question is if a bar plus a B is equals to 0 then we are ill wait to find values of a and B so see here we have been given with a bar plus a B that is equals to 0 so as if you apply distributive law you can see distributive law that we have already applied earlier right so if we apply distributed distributive law over here so in that case we can say this will be a bar plus B is equals to 0 now a bar plus B that is equals to 0 that is possible only if this is 0 and this is 0 so a bar is equals to 0 means a is equals to 1 and B is equals to 0 that situation results into a bar plus B is equals to 0 right so that is how even we can simply solve this type of question I hope that you have understood this question thank you so much for watching this video please do give your valuable suggestions definitely based on your suggestions in future I will make videos which will solve your queries thank you so much for watching this video